Chemistry

Calculate the [h30+] in a 0.010M solution of Sr(OH)2

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  1. 0.010M in Sr(OH)2.
    OH^- is twice that.
    (H3O^+)(OH^-) = Kw = 1E-14
    Solve for (H3O^+)

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  2. 5*10•14

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  3. because Sr(OH)2, do 0.010 x 2

    ans = 0.020

    pOH = -log(0.020)
    pOH = 1.7
    [finding pH because Sr(OH)2]
    pH = 14 - 1.7
    pH = 12.30

    10^-12.30 = 5.0x10^-13

    hope this helps

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